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I'm stuck with a problem regarding hybrid stepping motors and I'm not sure where I'm wrong.

I can't reconcile the torque constant with the maximum speed of a hybrid stepping motor with the following nameplate:

Holding torque = 9 Nm

Rated current = 10 A

Steps/rev = 200

Rated power = 40 W

The problem, in short, is this. The torque constant of this motor is 0.9 Nm/A, which (give or take some possible coefficients due to the multiphase nature of the machine or to the sinusoidal nature of the back-emf) is also the voltage constant in V/(rad/s) that yields the back-emf given the speed. The data sheet also gives the torque/speed characteristic: with a 24 V supply, the maximum speed is 6000 pps. With 200 steps/rev, 6000 pps is 30 rev/s, or 188 rad/s (mechanical). Now, if we multiply 188 rad/s times the voltage constant of 0.9, we get 169 V! How on Earth can a motor run on 24 V when its back-emf is 169 V? It should never be able to reach such speed.

I tried to write this off as a mistake in the data sheet, but then all the data sheets from this manufacturer must be wrong. Moreover, I found another data sheet from another manufacturer with the same problem (too high maximum speed given the supply voltage and machine constant).

What am I doing wrong? Thanks in advance to anyone who can shed some light.

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  • \$\begingroup\$ Sure you are not confusing \$K_T\$ with \$K_E\$? \$\endgroup\$ Commented Jun 21, 2020 at 17:16
  • \$\begingroup\$ Shouldn't they be almost the same? They are the same for DC motors; with multiphase machines you might have some corrective coefficients involved, but they won't differ by orders of magnitude. \$\endgroup\$
    – Carlo
    Commented Jun 21, 2020 at 17:25

2 Answers 2

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I think I found an answer. The torque constant and voltage constant are equal only in those machines in which torque generation is due only to the interaction between the armature and the permanent magnets, i.e. isotropic machines.

The share of electromagnetic torque due to anisotropy is proportional to the rate of change of inductance with angular position. Anisotropy will generate back-emfs as the rotor rotates only if an armature current is present; contrary to permanent magnet back-emf, anisotropy back-emf will not be detected measuring open-circuit phase voltages.

In machines having both means of torque generation, only the permanent magnet share of back-emf can be considered proportional to the mechanical speed through a voltage constant. In this case, the voltage constant is necessarily smaller than the torque constant, and is zero for fully anisotropic (i.e. reluctance) machines.

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Here is a hybrid stepper spec that shows them being equal, as expected:

Insulation class: Class B.

Dielectric strength: 500VDC for one minute.
Insulation resistance: 100MΩMin., 500VDC.

Electrical Specifications:
Model:  86BLS98 .
Number of poles: 8.
Number of phase: 3.
Rated voltage(DC): 48V.
Rated speed: 3000RPM.
Rated torque:  1.4N.m/197.2oz.in.
Rated Power: 440W.
Peak Torque: 4.2N.m/591.5oz.in.
Peak current: 33A.
Resistance: 0.2 ohms.
Inductance: 0.48mH±20%.
Torque constant: 0.13 Nm/A.
Back E.M.F: 13.5 V/KRPM.
Rotor inertia: 1600 g.cm2
Body length: 98mm/3.9in.

0.13Nm/A so \$K_T\$ = 0.13

13.5V/1000 RPM so 13.5V/16.66 rotations/s or 105 rad/s so \$K_E\$ = 0.13 .

Could be a mistake in their data I guess.

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  • \$\begingroup\$ Thanks, but the 86BLS98 is actually a DC brushless motor, not a hybrid stepper. (I looked it up because 8 poles 440 W sounded suspicious for a stepper). In this case the two constants are exactly the same (as for DC motors), as you found out. \$\endgroup\$
    – Carlo
    Commented Jun 21, 2020 at 18:33
  • \$\begingroup\$ Can you link the actual datasheet? \$\endgroup\$ Commented Jun 21, 2020 at 18:36
  • \$\begingroup\$ sonceboz.com/sites/default/files/medias/standard-products/… And here is another one that "suffers" from the same problem: ww1.microchip.com/downloads/en/DeviceDoc/… \$\endgroup\$
    – Carlo
    Commented Jun 21, 2020 at 18:39

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